Section 1.6

Homogeneous Systems, Subspaces, and Bases

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Homogeneous Systems

A homogeneous system of m equations with n unknowns x i ( i = 1.. n ) is a system of

the form

a[11]*x[1]+a[12]*x[2]+a[13]*x[3] + ........ + a[1*n]*x[n] = 0

a[21]*x[1]+a[22]*x[2]+a[23]*x[3] + ........ + a[2*n]*x[n] = 0

a[31]*x[1]+a[32]*x[2]+a[33]*x[3] + ........ + a[3*n]*x[n] = 0

.............

a[m1]*x[1]+a[m2]*x[2]+a[m3]*x[3] + ........ + a[mn]*x[n] = 0

Is a homogeneous system consistent or inconsistent?

Answer

The homogeneous system of linear equations is always consistent.

The solution set of a homogeneous system is never empty. In fact, x[1] = 0, x[2] = 0 , .. x[n] = 0 always satisfy such systems. This solution is called the zero or trivial solution .

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Example 1: Consider the system:

> eq1:=x[1]+x[2]+2*x[3]=0: eq2:=2*x[1]+x[2]-3*x[3]=0: print(eq1); print(eq2);

x[1]+x[2]+2*x[3] = 0

2*x[1]+x[2]-3*x[3] = 0

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What is the solution set of this system?

> solveqns({eq1,eq2},{x[1],x[2],x[3]});

Select one of the following modes

     1. Demonstration mode

     2. No-Step mode

2;

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`The values of the variables are given by`

vector([x[1] = 5*_t[1], x[2] = -7*_t[1], x[3] = _t[...

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Example 1 represents a system with infinitely many solutions and with one free variable. ******************************************************************

Example 2 Consider the system:

> eq1:=x[1]+2*x[2]-5*x[3]+6*x[4]=0: print(eq1);

x[1]+2*x[2]-5*x[3]+6*x[4] = 0

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What is the solution set of this system?

> solveqns(eq1,{x[1],x[2],x[3],x[4]});

Select one of the following modes

     1. Demonstration mode

     2. No-Step mode

2;

` `

`The values of the variables are given by`

vector([x[1] = -2*_t[1]+5*_t[2]-6*_t[3], x[2] = _t[...

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This system has infinitely many solutions with three free variable.

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Example 3: Consider the system:

> eq1:=x[1]-2*x[2]-x[3]=0: eq2:=2*x[1]-x[2]+x[3]=0: eq3:=x[1]+3*x[2]-5*x[3]=0:

> print(eq1); print(eq2); print(eq3);

x[1]-2*x[2]-x[3] = 0

2*x[1]-x[2]+x[3] = 0

x[1]+3*x[2]-5*x[3] = 0

What is the solution set of this system?

> solveqns({eq1,eq2,eq3},{x[1],x[2],x[3]});

Select one of the following modes

     1. Demonstration mode

     2. No-Step mode

2;

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`The values of the variables are given by`

vector([x[1] = 0, x[2] = 0, x[3] = 0])

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This system has the trivial solution as its unique solution.

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Examples illustrate several scenarios for homogeneous systems.

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If the number of equations is fewer than the number of unknowns ( under determined system ),

then the homogeneous system will have infinitely many solutions.

If the number of equations is greater or equal to the number of unknowns ( over determined

system ), then the homogeneous system will have either a unique solution or infinitely many solutions.

A homogeneous systems of linear equations is consistent. The solution set is either a

singleton or infinite.

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Every linear combination of solutions of a homogeneous system Ax=0 is again a solution of the system.

Nonhomogeneous Systems

Consider the nonhomogeneous system of m equations with n unknowns:

a[11]*x[1]+a[12]*x[2]+a[13]*x[3] + ........ + a[1*n]*x[n] = b[1]

a[21]*x[1]+a[22]*x[2]+a[23]*x[3] + ........ + a[2*n]*x[n] = b[2]

a[31]*x[1]+a[32]*x[2]+a[33]*x[3] + ........ + a[3*n]*x[n] = b[3]

.............

a[m1]*x[1]+a[m2]*x[2]+a[m3]*x[3] + ........ + a[mn]*x[n] = b[m]

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Every system of linear equations Ax = b has either no solution, exactly one solution or

infinitely many solutions.

Example 4: Consider the system:

> eq1:=x[1]+x[2]+2*x[3]=1: eq2:=2*x[1]+2*x[2]-4*x[3]=2:

> print(eq1); print(eq2);

x[1]+x[2]+2*x[3] = 1

2*x[1]+2*x[2]-4*x[3] = 2

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What is the solution set of this system?

> solveqns({eq1,eq2},{x[1],x[2],x[3]});

Select one of the following modes

     1. Demonstration mode

     2. No-Step mode

2;

` `

`The values of the variables are given by`

vector([x[1] = 1-_t[1], x[2] = _t[1], x[3] = 0])

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This system has infinitely many solutions with one free variable. The solution of the

nonhomogeneous system can be expressed as

> matrix(3,1,[x[1],x[2],x[3]]) = matrix(3,1,[1,0,0]) + a*matrix(3,1,[-1,1,0]);

matrix([[x[1]], [x[2]], [x[3]]]) = matrix([[1], [0]...

where x[2] = a is the free variable.The solution [1,0,0] is called the particular solution to the system; while a*[-1, 1, 0] is a solution of the associated homogeneous system.

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Example 5: Consider the system:

> eq1:=x[1]+2*x[2]-5*x[3]+6*x[4]=10: print(eq1);

x[1]+2*x[2]-5*x[3]+6*x[4] = 10

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What is the solution set of this system?

> solveqns(eq1,{x[1],x[2],x[3],x[4]});

Select one of the following modes

     1. Demonstration mode

     2. No-Step mode

2;

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`The values of the variables are given by`

vector([x[1] = 10-2*_t[1]+5*_t[2]-6*_t[3], x[2] = _...

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This system has infinitely many solutions with three free variables . The general solution

[ x[1], x[2], x[3], x[4] ] of the nonhomogeneous system is

[x[1], x[2], x[3], x[4]] = [10-2*a+5*b-6*c, a, b, c...

= [10, 0, 0, 0]+[-2*a, a, 0, 0]+[5*b, 0, b, 0]+[-6*c,...

= [10, 0, 0, 0]+a*[-2, 1, 0, 0]+b*[5, 0, 1, 0]+c*[-6,...

The solution [10,0,0,0] is the particular solution of the system; while the solutions

a*[-2, 1, 0, 0] ; b*[5, 0, 1, 0] ; and c*[-6, 0, 0, 1] are solutions to the associated

homogeneous system.

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Let Ax=b be a linear system. If p is any particular solution of Ax=b and h is a solution of the

corresponding homogeneous system Ax=0 , then p+h is a solution of Ax=b . Moreover, every

solution of Ax=b has tghe form p+h, so that the general solution is x=p+h where Ah=0 .

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Subspaces

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A subset W of R^n is closed under vector addition if for all u,v in W the sum u+v is in W .

If r v is in W for all v in W and all scalars r, then W is closed under scalar multiplication.

A non-empty subset W of R^n that is closed under both vector addition and scalar

multiplication is a subspace of R^n .

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Example 6:

> `S1 ={[x,y,z] | 2*x+3*y+z=10}`;

`S1 ={[x,y,z] | 2*x+3*y+z=10}`

Choose two vectors in S[1] :

> v:=vector([-4,3,9]);u:=vector([-7,4,12]);

v := vector([-4, 3, 9])

u := vector([-7, 4, 12])

Compute their sum

> `v+u`=evalm(v+u);

`v+u` = vector([-11, 7, 21])

Does vector v + u belong to subset S[1] ? Why?

Example 7:

> `S2 ={[x,y,z] | 2*x+3*y+z=0}`;

`S2 ={[x,y,z] | 2*x+3*y+z=0}`

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Is subset S[2] closed under addition? under scalar multiplication?

Answer

Subset S[2] is closed under the two operations. Hence S[2] is a R^2 subspace of R^2 .

> with(plots):

Warning, the name changecoords has been redefined

> implicitplot3d({2*x+3*y+z=0,2*x+3*y+z=10},x=-5..5,y=-5..5,z=0..10,axes=normal);

[Maple Plot]

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We observe that the graph of S[2] passes through the origin while the graph of S[1] does not.

The zero vector belongs to the set S[2] but does not belong to the set S[1] .

Graphically, can you tell which subset does not contain the zero vector?

What is the implication of not having the zero vector in the set?

Example 8: Consider the following set

> `S1 = {[x,y] | x^2 + y^2 <= 1}`;

`S1 = {[x,y] | x^2 + y^2 <= 1}`

Note that the set {0} and the set R^n are subspaces of R^n .

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Let W = sp{ w[1], w[2], w[3] , . . ., w[k] } be the span of k vectors in R^n . Then W is a subspace

of R^n .

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Example 9: Consider the homogeneous system of linear equations

> eq1:=x1 + 3*x2 + 5*x3 = 0:
eq2:=3*x1+ 2*x2 + 5*x3 = 0:

> print(eq1); print(eq2);

x1+3*x2+5*x3 = 0

3*x1+2*x2+5*x3 = 0

The system is defined by matrix multiplication A*x = 0 where A and x are

> A:=matrix([[1,3,5],[3,2,5]]);
x:=matrix([[x1],[x2],[x3]]);

A := matrix([[1, 3, 5], [3, 2, 5]])

x := matrix([[x1], [x2], [x3]])

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which is equivalent to

> multiply(A,x)=0;

matrix([[x1+3*x2+5*x3], [3*x1+2*x2+5*x3]]) = 0

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Upon solving the system

> solve({eq1,eq2},{x1,x2,x3});

{x2 = 2*x1, x3 = -7/5*x1, x1 = x1}

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we obtain

> `S ={[x1,x2,x3]|x1=(-5/7)*a,x2=(-10/7)*a,x3=a}`;

`S ={[x1,x2,x3]|x1=(-5/7)*a,x2=(-10/7)*a,x3=a}`

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or

> S = span(matrix(3,1,[-5/7,-10/7,1]));

S = span(matrix([[-5/7], [-10/7], [1]]))

or

> S = span(matrix(3,1,[-5,-10,7]));

S = span(matrix([[-5], [-10], [7]]))

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Is S a subspace of R^3 ?

Geometrically, S represents a line through the origin. The set S is closed under addition and

scalar multiplication. Therefore, S is a subspace. Hence, the solution set of a homogeneous

system is a subspace. This is called the nullspace of the matrix A and is denoted by N(A).

Is the solution set of a nonhomogeneous system of linear equations a subspace?

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The ROW SPACE of an nxm matrix A is the space spanned by the rows of the matrix A.

Since the span of a set of vectors is a subspace, the row space of an nxm matrix A is a

subspace of R^m .

The COLUMN SPACE of an nxm matrix A is the space spanned by the columns of the

matrix A.

Since the span of a set of vectors is a subspace, the column space of an nxm matrix A is a

The NULL SPACE of an nxm matrix A is the solution set to Ax=0.

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Example 10: Find the row space, column space, and the null space.

> A:=matrix([[1,8,3],[6,-3,7]]);

A := matrix([[1, 8, 3], [6, -3, 7]])

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> AUG := matrix([[1,8,3,0],[6,-3,7,0]]);

AUG := matrix([[1, 8, 3, 0], [6, -3, 7, 0]])

> solve({x1+8*x2+3*x3=0,6*x1-3*x2+7*x3=0},{x1,x2,x3});

{x1 = x1, x2 = 11/65*x1, x3 = -51/65*x1}

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A linear system Ax=b has a solution if and only if b is in the column space of A.

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Bases

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Let W be a subpsace of R^n . A subset { w[1], w[2], w[3] ,. . ., w[k] } of W is a basis for W if

every vector in W can be expressed uniquely as a linear combination of w[1], w[2], w[3] ,. . ., w[k] .

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Example 11: Consider the two independent vectors

> e1:=vector([1,0]); e2:=vector([0,1]);

e1 := vector([1, 0])

e2 := vector([0, 1])

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Is every vector in R^2 a linear combination of the vectors e[1] and e[2] ?

Let w be any vector in R^2

> w:= vector([x,y]);

w := vector([x, y])

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If w = [x,y] is a linear combination of e[1] and e[2]

> evalm(w)=c1*evalm(e1)+c2*evalm(e2);

vector([x, y]) = c1*vector([1, 0])+c2*vector([0, 1]...

Then c[1] = x and c[2] = y. Hence w is

> evalm(w)=x*evalm(e1)+y*evalm(e2);

vector([x, y]) = x*vector([1, 0])+y*vector([0, 1])

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Therefore every vector in R^2 can be written uniquely as a linear combination of the two

vectors e[1] and e[2] . Thus the set { e[1] , e[2] } is spans R^2 and is a basis for R^2 .

Example 12: Now consider the vectors

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> e1:=vector([1,0]);e2:=vector([0,1]);
u:=vector([2,5]);

e1 := vector([1, 0])

e2 := vector([0, 1])

u := vector([2, 5])

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Is the set { e[1] , e[2] , u} a basis for R^2 ? Let w be any vector in R^2

> w:=vector([x,y]);

w := vector([x, y])

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If w = [x , y] is a linear combination of e[1] , e[2] and u

> evalm(w)=c1*evalm(e1)+c2*evalm(e2)+c3*evalm(u);

vector([x, y]) = c1*vector([1, 0])+c2*vector([0, 1]...

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This yields the system of equations

> eq1:=c1+2*c3=x; eq2:=c2+5*c3=y;

eq1 := c1+2*c3 = x

eq2 := c2+5*c3 = y

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On solving the system we get

> solve({eq1,eq2},{c1,c2,c3});

{c2 = -5*c3+y, c1 = -2*c3+x, c3 = c3}

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Since the system is consistent, the vectors e[1] , e[2] , and u do span R^2 . However,

the linear combination is not unique.

> evalm(w)=(x-2)*evalm(e1)+(y-5)*evalm(e2)+evalm(u);

vector([x, y]) = (x-2)*vector([1, 0])+(y-5)*vector(...

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The set { w[1], w[2], w[3] ,. . . , w[k] } is a basis for W = sp{ w[1], w[2], w[3] ,. . . , w[k] } in

R^n if and only if the zero vector is a unique linear combination of the w[i] that is

if and only if r[1]*w[1]+r[2]*w[2]+r[3]*w[3] + . . . + r[k]*w[k] = 0 implies that

r[1] = r[2] = r[3] = r[4] = . . . = r[k] = 0.

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We call the set { e[1], e[2], e[3] , . . . , e[n] } the standard basis for R^n .

Example 13: Consider the vectors

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> v1:=vector([0,1,1]); v2:=vector([1,1,2]);v3:=vector([1,0,2]);

v1 := vector([0, 1, 1])

v2 := vector([1, 1, 2])

v3 := vector([1, 0, 2])

> eq1:=c2 + c3=0: eq2:=c1+ c2=0: eq3:= c1+2*c2+2*c3=0:

> print(eq1); print(eq2); print(eq3);

c2+c3 = 0

c1+c2 = 0

c1+2*c2+2*c3 = 0

> solve({eq1,eq2,eq3},{c1,c2,c3});

{c3 = 0, c1 = 0, c2 = 0}

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Change vector v1

> v1 := vector([3,-2,6]);

v1 := vector([3, -2, 6])

> eq1:= 3*c1 + c2 + c3=0: eq2:=-2*c1+ c2=0: eq3:= 6*c1+2*c2+2*c3=0:

> print(eq1); print(eq2); print(eq3);

3*c1+c2+c3 = 0

-2*c1+c2 = 0

6*c1+2*c2+2*c3 = 0

> solve({eq1,eq2,eq3},{c1,c2,c3});

{c3 = -5/2*c2, c2 = c2, c1 = 1/2*c2}

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Unique Solutions for Ax=b

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Let A be a square matrix, i.e. A is a n x n matrix. The following are equivalent:

1.) The linear system Ax=b has a unique solution

2.) The matrix A is row equivalent to the identity matrix I.

3.) The matrix A is invertible.

4.) The column vectors of A form a basis for R^n

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Let A be an m x n matrix. The following are equivalent.

1.) Each consistent system Ax=b has a unique solution

2.) The reduced row-echelon form of A consists of the n x n identity matrix

followed by m-n rows of zeros

3.) The columns of A form a basis for the column space A .

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Exercises

1, 3, 9, 12, 14, 17, 19, 23, 25, 29, 30,31, 38.